3.358 \(\int \frac{a+b \sin ^{-1}(c x^2)}{x^6} \, dx\)

Optimal. Leaf size=61 \[ \frac{2}{15} b c^{5/2} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{c} x\right ),-1\right )-\frac{a+b \sin ^{-1}\left (c x^2\right )}{5 x^5}-\frac{2 b c \sqrt{1-c^2 x^4}}{15 x^3} \]

[Out]

(-2*b*c*Sqrt[1 - c^2*x^4])/(15*x^3) - (a + b*ArcSin[c*x^2])/(5*x^5) + (2*b*c^(5/2)*EllipticF[ArcSin[Sqrt[c]*x]
, -1])/15

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Rubi [A]  time = 0.0338484, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4842, 12, 325, 221} \[ -\frac{a+b \sin ^{-1}\left (c x^2\right )}{5 x^5}-\frac{2 b c \sqrt{1-c^2 x^4}}{15 x^3}+\frac{2}{15} b c^{5/2} F\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x^2])/x^6,x]

[Out]

(-2*b*c*Sqrt[1 - c^2*x^4])/(15*x^3) - (a + b*ArcSin[c*x^2])/(5*x^5) + (2*b*c^(5/2)*EllipticF[ArcSin[Sqrt[c]*x]
, -1])/15

Rule 4842

Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSin[
u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/Sqrt[1 - u^2], x]
, x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(
m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rubi steps

\begin{align*} \int \frac{a+b \sin ^{-1}\left (c x^2\right )}{x^6} \, dx &=-\frac{a+b \sin ^{-1}\left (c x^2\right )}{5 x^5}+\frac{1}{5} b \int \frac{2 c}{x^4 \sqrt{1-c^2 x^4}} \, dx\\ &=-\frac{a+b \sin ^{-1}\left (c x^2\right )}{5 x^5}+\frac{1}{5} (2 b c) \int \frac{1}{x^4 \sqrt{1-c^2 x^4}} \, dx\\ &=-\frac{2 b c \sqrt{1-c^2 x^4}}{15 x^3}-\frac{a+b \sin ^{-1}\left (c x^2\right )}{5 x^5}+\frac{1}{15} \left (2 b c^3\right ) \int \frac{1}{\sqrt{1-c^2 x^4}} \, dx\\ &=-\frac{2 b c \sqrt{1-c^2 x^4}}{15 x^3}-\frac{a+b \sin ^{-1}\left (c x^2\right )}{5 x^5}+\frac{2}{15} b c^{5/2} F\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )\\ \end{align*}

Mathematica [C]  time = 0.133281, size = 72, normalized size = 1.18 \[ -\frac{-2 i b (-c)^{5/2} x^5 \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-c} x\right ),-1\right )+3 a+2 b c x^2 \sqrt{1-c^2 x^4}+3 b \sin ^{-1}\left (c x^2\right )}{15 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c*x^2])/x^6,x]

[Out]

-(3*a + 2*b*c*x^2*Sqrt[1 - c^2*x^4] + 3*b*ArcSin[c*x^2] - (2*I)*b*(-c)^(5/2)*x^5*EllipticF[I*ArcSinh[Sqrt[-c]*
x], -1])/(15*x^5)

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Maple [A]  time = 0.01, size = 87, normalized size = 1.4 \begin{align*} -{\frac{a}{5\,{x}^{5}}}+b \left ( -{\frac{\arcsin \left ( c{x}^{2} \right ) }{5\,{x}^{5}}}+{\frac{2\,c}{5} \left ( -{\frac{1}{3\,{x}^{3}}\sqrt{-{c}^{2}{x}^{4}+1}}+{\frac{1}{3}{c}^{{\frac{3}{2}}}\sqrt{-c{x}^{2}+1}\sqrt{c{x}^{2}+1}{\it EllipticF} \left ( x\sqrt{c},i \right ){\frac{1}{\sqrt{-{c}^{2}{x}^{4}+1}}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x^2))/x^6,x)

[Out]

-1/5*a/x^5+b*(-1/5/x^5*arcsin(c*x^2)+2/5*c*(-1/3*(-c^2*x^4+1)^(1/2)/x^3+1/3*c^(3/2)*(-c*x^2+1)^(1/2)*(c*x^2+1)
^(1/2)/(-c^2*x^4+1)^(1/2)*EllipticF(x*c^(1/2),I)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \arcsin \left (c x^{2}\right ) + a}{x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^6,x, algorithm="fricas")

[Out]

integral((b*arcsin(c*x^2) + a)/x^6, x)

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Sympy [A]  time = 3.44533, size = 61, normalized size = 1. \begin{align*} - \frac{a}{5 x^{5}} + \frac{b c \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{1}{2} \\ \frac{1}{4} \end{matrix}\middle |{c^{2} x^{4} e^{2 i \pi }} \right )}}{10 x^{3} \Gamma \left (\frac{1}{4}\right )} - \frac{b \operatorname{asin}{\left (c x^{2} \right )}}{5 x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x**2))/x**6,x)

[Out]

-a/(5*x**5) + b*c*gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), c**2*x**4*exp_polar(2*I*pi))/(10*x**3*gamma(1/4)) - b
*asin(c*x**2)/(5*x**5)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arcsin \left (c x^{2}\right ) + a}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x^2))/x^6,x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x^2) + a)/x^6, x)